Download Circuits Lecture 3: Superposition

Circuits
Lecture 4: Superposition
李宏毅 Hung-yi Lee
Outline
• Matrix Equation for Node and Mesh analysis
• Chapter 4.1, 4.2
• Superposition
• Chapter 2.4
Node Analysis
vs  v1  v1 v2  v1


 is  0
v1:
Ra
Rb
Rc
v v v v v
v2: 1 2  2  3 2  0
Rc
Rd
Re
v3:
vs
 v1  v1 v2  v1


 is 
Ra
Rb
Rc
Ra
v1  v2  v2 v3  v2


0
Rc
Rd
Re
v2  v3  v3

 is
Re
Rf
v2  v3  v3

 is
Re
Rf
 1
v
1
1 
1
 

 v1  v2  is  s
Rc
Ra
 Ra Rb Rc 
 1
1
1
1 
1
v1    
 v2  v3  0
Rc
Re
 Rc Rd Re 
 1
1
1 

v2   
v3  is


Re
 Re R f 
Node Analysis
 1
1
1




 Ra Rb Rc
1


Rc


0

1
Rc
1
1
1
 

Rc Rd Re
1
Re
Rv  s 
Resistance
Node
Sources
potentials
You can directly write the
matrix equation below.
(textbook, P139)

vs 
0


  v1   is  R 
a
1

  
v

0

 2  
Re

 v3   is
1
1


  


Re R f 
 1
v
1
1 
1
 

 v1  v2  is  s
Rc
Ra
 Ra Rb Rc 
 1
1
1
1 
1
v1    
 v2  v3  0
Rc
Re
 Rc Rd Re 
 1
1
1 

v2   
v3  is


Re
 Re R f 
Node Analysis
Rv  s 
v  R s 
1
 1
1
1



 Ra Rb Rc
1


Rc


0

1
Rc
1
1
1
 

Rc Rd Re
1
Re
 v1   a11
v    a
 2   21
v3  a31
a12
a22
a32

vs 



i

  v1   s R 
a
1

  
v

0
2



 
Re


 
1
1  v3   is

 


Re R f 
0
vs 

a13   is  R 
a



a23   0 

a33   is




v1, v2, v3 is the weighted sum of is and vs
Node potential is the weighted sum of the values
of sources
Voltage (potential difference) is the weighted
sum of the values of sources
Mesh Analysis
For mesh 1
Ra(i1-is)+Rbi1+Rc(i1-i2)-vs=0
For mesh 2
Rc(i2-i1)+Rdi2+Re(i2-i3)=0
For mesh 3
Re(i3-i2)+Rfi3+vs=0
 Ra  Rb  Rc

 Rc


0

 Rc
Rc  Rd  Re
 Re
You can directly write the
matrix equation below.
0   i1   Ra is  vs 
  
 Re  i2    0 
Re  R f  i3    vs 
(textbook, P153)
Mesh Analysis
Resistance
Mesh
Current
Sources
Ri   s 
 Ra  Rb  Rc

 Rc


0

 Rc
Rc  Rd  Re
 Re
0   i1   Ra is  vs 
  
 Re  i2    0 
Re  R f  i3    vs 
Mesh Analysis
Ri   s 
i   R s 
1
 Ra  Rb  Rc

 Rc


0

 Rc
Rc  Rd  Re
 Re
 i1  b11 b12
i   b
 2   21 b22
i3  b31 b32
0   i1   Ra is  vs 

 Re  i2    0 
Re  R f  i3    vs 
b13   Ra is  vs 
b23   0 
b33    vs 
i1, i2, i3 is the weighted sum of is and vs
Mesh currents are the weighted sum of the values
of sources
Currents of the braches are the weighted sum of
the values of sources
Linearity
Based on node and mesh analysis:
y   ai xi
i
• y: any current or voltage for an element
• xi: current of current sources or voltage of
voltage sources
Any current (or voltage) for an element is the
weighted sum of the voltage (or current) of the
sources.
Linearity - Example
i1  a1 x1  a2 x2  a3 x3
x1
x2
x3
Any current (or voltage) for an element is the
weighted sum of the voltage (or current) of the
sources.
Not apply on Power
• xi: current of independent current sources or voltage of
independent voltage sources
Voltage:
v   ai xi
i
Power:
Current:
i   bi xi
i
p   ci xi
i
Power:



p  vi    ai xi   bi xi 
 i
 i

Proportionality Principle – One
Independent Sources
Find i1 and v1
when vs is 9V,
72V and 0.9V
Complex
Circuit
i1  a1vs
v1  a2 vs
vs  9V
i1  1A
v1  10V
vs  72V
i1  8 A
v1  80V
vs  0.9V i1  0.1A
v1  1V
Superposition Principle – Multiple
Independent Sources
• Example 2.10
• Find i1
x2
i1  a1 x1  a2 x2  a3 x3
 i11  i1 2  i13
We can find i1-1, i1-2, i1-3
separately.
x1
x3
When x2=0 and x3=0,
The current through 2Ω is i1-1.
Superposition Principle – Multiple
Independent Sources
x2  0
• Example 2.10
• Find i1
i1  a1 x1  a2 x2  a3 x3
 i11  i1 2  i13
We can find i1-1, i1-2, i1-3
separately.
Current of current
source set to be zero.
x1
x3  0
Open
Circuit
i11  30 / 6  4  2   2.5 A
Superposition Principle – Multiple
Independent Sources
• Example 2.10
• Find i1
x2
i1  a1 x1  a2 x2  a3 x3
 i11  i1 2  i13
We can find i1-1, i1-2, i1-3
separately.
x1
x3
To find i1-2, we set x1=0 and x3=0.
Now the current through 2Ω is i1-2.
Superposition Principle – Multiple
Independent Sources
x2
• Example 2.10
• Find i1
i1  a1 x1  a2 x2  a3 x3
 i11  i1 2  i13
We can find i1-1, i1-2, i1-3
separately.
x1  0
Voltage of voltage
source set to be zero.
x3  0
Short
Circuit
i1 2  3  4 / 2  6   4   1A
Superposition Principle – Multiple
Independent Sources
x 0
2
• Example 2.10
• Find i1
i1  a1 x1  a2 x2  a3 x3
 i11  i1 2  i13
We can find i1-1, i1-2, i1-3
separately.
i13
x1  0
x3
6
  8
 4 A
6  4  2 
Superposition Principle – Multiple
Independent Sources
• Example 2.10
• Find i1
x2
i1  a1 x1  a2 x2  a3 x3
 i11  i1 2  i13
We can find i1-1, i1-2, i1-3
separately.
set x2=0 and x3=0
set x1=0 and x3=0
set x1=0 and x2=0
x1
i11  2.5 A
i1 2  1A
i13  4 A
x3
i1  0.5 A
Superposition Principle – Multiple
Independent Sources
• Steps to apply Superposition Principle:
• If the circuit has multiple sources, to find a voltage or
current for an element
• For each source
• Keep the source unchanged
• All the other sources set to zero
• Voltage source’s voltage set to 0 = Short circuit
• Current source’s current set to 0 = open circuit
• Find the voltage or current for the element
• Add all the voltages or currents obtain by individual
sources
Remind
• Always using superposition when there are
multiple sources?
One circuit (3 sources)
Three circuits (1 source)
v.s.
Concluding Remarks
y   ai xi
This equation only for circuits
with sources and resistors.
i
•
•
y: any current or voltage for an element
xi: current of current sources or voltage of voltage sources
Proportionality Principle, Superposition Principle
Can be used in any circuit in this course
Linearity
• A circuit is a multiple-input multiple-output (MIMO)
system
• Input: current of current sources or voltage of voltage
sources
• Output: the current or voltage for the elements
input
Circuit
(System)
+
v
-
i
output
All circuits in this
courses are
linear circuits.
Linearity
• All linear circuits are linear system
• Linear Circuit:
All circuits in this
• Sources
courses are
• Linear Elements:
linear
systems.
• Resistor, Capacitor, Inductor
v
i
R
Linearity
• Linear System:
• Property 1:
Input: g1(t), g2(t), g3(t), ……
output: h1(t), h2(t), h3(t), ……
Input: Kg1(t), Kg2(t), Kg3(t), ……
output: Kh1(t), Kh2(t), Kh3(t), ……
Proportionality Principle
Linearity
• Linear System:
• Property 2:
Input: a1(t), a2(t), a3(t), ……
output: x1(t), x2(t), x3(t), ……
Input: b1(t), b2(t), b3(t), ……
output: y1(t), y2(t), y3(t), ……
Input: a1(t)+ b1(t), a2(t)+ b2(t), a3(t)+ b3(t), ……
output: x1(t)+y1(t), x2(t)+y2(t), x3(t)+y3(t), ……
Superposition Principle
Linearity
• Linear System:
• Property 2:
Input: a1(t), a2(t), a3(t), ……
output: x1(t), x2(t), x3(t), ……
Input: b1(t), b2(t), b3(t), ……
output: y1(t), y2(t), y3(t), ……
Input: a1(t)+ b1(t), a2(t)+ b2(t), a3(t)+ b3(t), ……
output: x1(t)+y1(t), x2(t)+y2(t), x3(t)+y3(t), ……
Superposition Principle
vt 
Linearity
vt 
i t 
g1 t 
0
g t 
g t   g1 t   g 2 t 
Superposition Principle can be
applied on any circuit in this
course (Textbook: Chapter 6.5).
0
i t 
g 2 t 
Homework
• 2.50
Given vs
and R3,
find vb
Homework
• 2.52
Given is, find vs such that v4= 36V
Thank you!
Answer
• 2.50
• -12V
• 2.52
• 60V